If sides A and B of a triangle have lengths of 13 and 2 respectively, and the angle between them is $\frac{π}{6}$ $(pi)/6$ , then what is the area of the triangle?

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If sides A and B of a triangle have lengths of 13 and 2 respectively, and the angle between them is $\frac{π}{6}$ $(pi)/6$ , then what is the area of the triangle?

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Answer 1 · 2016-02-06T06:04:45

First we solve $\frac{π}{6}$ $pi/6$
$π = 180^{\circ}$ $pi=180^circ$

So,

$\to = \frac{180}{6} = 45^{\circ}$ $rarr=180/6=45^circ$

Now we consider the diagram:

enter image source here

The formula for finding the area of triangle when given two sides of a triangle and the angle between them= $\frac{1}{2} a b sin C = \frac{a b sin C}{2}$ $1/2 ab sinC=(ab sinC)/2$

In this case $a = 13, b = 2, C = 45^{\circ}$ $a=13,b=2,C=45^circ$

$sin C = sin (45^{\circ}) = \frac{\sqrt{2}}{2}$ $sin C=sin(45^circ)=sqrt2/2$

$\to A r e a = \frac{(13) (2) (\frac{\sqrt{2}}{2})}{2}$ $rarrArea=((13)(2)(sqrt2/2))/2$

$\to A r e a = \frac{26 (\frac{\sqrt{2}}{2})}{2}$ $rarrArea=(26(sqrt2/2))/2$

$\to A r e a = \frac{\frac{52 \sqrt{2}}{2}}{2}$ $rarrArea=((52sqrt2)/2)/2$

$\to A r e a = \frac{26 \sqrt{2}}{2} = 13 \sqrt{2}$ $rarrArea=(26sqrt2)/2=13sqrt2$

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If sides A and B of a triangle have lengths of 13 and 2 respectively, and the angle between them is $\frac{π}{6}$ $(pi)/6$ , then what is the area of the triangle?

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If sides A and B of a triangle have lengths of 13 and 2 respectively, and the angle between them is π6(pi)/6, then what is the area of the triangle?

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If sides A and B of a triangle have lengths of 13 and 2 respectively, and the angle between them is $\frac{π}{6}$ $(pi)/6$ , then what is the area of the triangle?